CalcGospel 國際數學圖譜
返回

IAL 2022 June Q7

A Level / Edexcel / P3

IAL 2022 June Paper · Question 7

题目

Problem

In this question you must show all stages of your working.

Solutions relying entirely on calculator technology are not acceptable.

(a) Show that the equation

2sin⁡θ(3cot⁡22θ−7)=13sec⁡θ2\sin\theta(3\cot^2 2\theta-7)=13\sec\theta

can be written as

3cosec⁡22θ−13cosec⁡2θ−10=03\cosec^2 2\theta-13\cosec 2\theta-10=0
(4)

(b) Hence solve, for 0<θ<π20<\theta<\dfrac{\pi}{2}, the equation

2sin⁡θ(3cot⁡22θ−7)=13sec⁡θ2\sin\theta(3\cot^2 2\theta-7)=13\sec\theta

giving your answers to 3 significant figures.

(4)
题目中文翻译

本题中你必须写出解题过程的所有步骤。

不接受完全依赖计算器技术的解法。

(a) 证明方程

2sin⁡θ(3cot⁡22θ−7)=13sec⁡θ2\sin\theta(3\cot^2 2\theta-7)=13\sec\theta

可写成

3cosec⁡22θ−13cosec⁡2θ−10=03\cosec^2 2\theta-13\cosec 2\theta-10=0

的形式。

(b) 由此在 0<θ<π20<\theta<\dfrac{\pi}{2} 内解方程

2sin⁡θ(3cot⁡22θ−7)=13sec⁡θ2\sin\theta(3\cot^2 2\theta-7)=13\sec\theta

答案保留 3 位有效数字。

解答

(a)

Start with

2sin⁡θ(3cot⁡22θ−7)=13sec⁡θ2\sin\theta(3\cot^2 2\theta-7)=13\sec\theta

Divide both sides by 2sin⁡θ2\sin\theta:

3cot⁡22θ−7=13sec⁡θ2sin⁡θ3\cot^2 2\theta-7 =\frac{13\sec\theta}{2\sin\theta}

Since

sec⁡θ=1cos⁡θ\sec\theta=\frac1{\cos\theta}

the right hand side becomes

132sin⁡θcos⁡θ\frac{13}{2\sin\theta\cos\theta}

Using

sin⁡2θ=2sin⁡θcos⁡θ\sin2\theta=2\sin\theta\cos\theta

we get

3cot⁡22θ−7=13cosec⁡2θ3\cot^2 2\theta-7 =13\cosec2\theta

Now use

cot⁡22θ=cosec⁡22θ−1\cot^2 2\theta=\cosec^2 2\theta-1

So

3(cosec⁡22θ−1)−7=13cosec⁡2θ3(\cosec^2 2\theta-1)-7=13\cosec2\theta

Expand:

3cosec⁡22θ−3−7=13cosec⁡2θ3\cosec^2 2\theta-3-7=13\cosec2\theta

Thus

3cosec⁡22θ−10=13cosec⁡2θ3\cosec^2 2\theta-10=13\cosec2\theta

Therefore

3cosec⁡22θ−13cosec⁡2θ−10=0\boxed{3\cosec^2 2\theta-13\cosec2\theta-10=0}

as required.

(b)

From part (a),

3cosec⁡22θ−13cosec⁡2θ−10=03\cosec^2 2\theta-13\cosec2\theta-10=0

Let

u=cosec⁡2θu=\cosec2\theta

Then

3u2−13u−10=03u^2-13u-10=0

Factorise:

(3u+2)(u−5)=0(3u+2)(u-5)=0

So

u=−23u=-\frac23

or

u=5u=5

Since ∣cosec⁡2θ∣≥1|\cosec2\theta|\ge1, reject

u=−23u=-\frac23

Thus

cosec⁡2θ=5\cosec2\theta=5

and hence

sin⁡2θ=15\sin2\theta=\frac15

For

0<θ<π20<\theta<\frac{\pi}{2}

we have

0<2θ<π0<2\theta<\pi

So

2θ=sin⁡−1152\theta=\sin^{-1}\frac15

or

2θ=π−sin⁡−1152\theta=\pi-\sin^{-1}\frac15

Therefore

θ=12sin⁡−115\theta=\frac12\sin^{-1}\frac15

or

θ=12(π−sin⁡−115)\theta=\frac12\left(\pi-\sin^{-1}\frac15\right)

Hence

θ=0.1006…, 1.470…\theta=0.1006\ldots,\ 1.470\ldots

So, to 3 significant figures,

θ=0.101, 1.47\boxed{\theta=0.101,\ 1.47}